Question

方程组x+y-a=0,5x+3y=31有正数解,确定a的取值范围

Answer 1

x+y-a=0 (1)

5x+3y=31 (2)

(1)×5-(2)

5y-5a-3y=0-32

2y=5a-32

y=(5a-32)/2>0

5a-32>0

a>32/5

代入(1)

x=a-y=a-(5a-32)/2>0

(5a-32)/2